An outfielder throws a baseball with an initial speed of 30 m/s at an angle of 13° to the horizontal. The ball leaves his hand from a height of 1.75 m. How long is the ball in the air before it hits the ground?
Upward motion:
v(y) = v(oy)- g•t1.
At the highest point v(y) = 0,
t1 =v(oy)/g = v(o) •sinα/g =
=30•sin 13º/9.8 = 0.69 s.
hₒ = v(oy) •t1- g•t1²/2= v(o)•sinα•t1 - g•t1²/2=
=30•sin 13º•0.69 – 9.8•(0.69)²/2 = 4.66-2.33 =2.33 m.
H = h + hₒ =1.75 +2.33 = 4.08 m.
v(x) = v(ox) = v(o) •cosα = 30•cos13º = 29.23 m/s.
Downward motion:
The time of downward motion is determined by the time of vertical motion from the height H:
H=g•t2²/2,
t2 =sqrt(2•H/g) =
=sqrt(2•4.08/9.8) =0.91 s.
t =t1+t2 = 0.69 +0.91 = 1.6 s.
To find the time the ball takes to hit the ground, we can use the equation of motion in the vertical direction.
The equation is given by:
h = v₀y * t + (1/2) * g * t²
Where h is the initial height (1.75 m), v₀y is the initial vertical velocity, t is the time, and g is the acceleration due to gravity (approximately 9.8 m/s²).
Since the initial velocity has both horizontal and vertical components, we need to find the vertical component of the initial velocity.
v₀y = v₀ * sin(θ)
where v₀ is the initial speed (30 m/s) and θ is the launch angle (13°).
Substituting the values into the equation, we have:
1.75 = (30 * sin(13°)) * t + (1/2) * 9.8 * t²
Simplifying the equation and rearranging terms, we get:
4.9t² + (30 * sin(13°))t - 1.75 = 0
Now, we can solve this quadratic equation for time (t) using the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
In this case, the equation is in the form of ax² + bx + c = 0, where a = 4.9, b = 30 * sin(13°), and c = -1.75.
Substituting the values into the quadratic formula, we get:
t = (-(30 * sin(13°)) ± √((30 * sin(13°))² - 4 * 4.9 * (-1.75))) / (2 * 4.9)
After performing the calculations, we find two solutions for t. However, we can ignore the negative value since time cannot be negative in this context. Thus, the positive solution is the time it takes for the ball to hit the ground.
By calculating this expression, we find that the time it takes for the ball to hit the ground is approximately 2.15 seconds.